1. Optical Fibre Dispersion By: Mr. Gaurav Verma Asst. Prof. ECE NIEC

2. Why does dispersion matter ? Understanding the effects of dispersion in optical fibers is quite essential in optical communications in order to minimize pulse spreading. Pulse compression due to negative dispersion can be used to shorten pulse duration in chirped pulse lasers

45. Dispersion in Multimode Step Index Fiber θaθa Easy Derivation from Senior or Sapna Katyar…

53. Birefringence in single-mode fibers Because of asymmetries the refractive indices for the two degenerate modes (vertical & horizontal polarizations) are different. This difference is referred to as birefringence, : Optical Fiber communications, 3 rd ed.,G.Keiser,McGrawHill, 2000

54. Fiber Beat Length In general, a linearly polarized mode is a combination of both of the degenerate modes. As the modal wave travels along the fiber, the difference in the refractive indices would change the phase difference between these two components & thereby the state of the polarization of the mode. However after certain length referred to as fiber beat length, the modal wave will produce its original state of polarization. This length is simply given by: [2-35]

55. Modal Birefringence

56. Intermodal Dispersion

57.  Exists in multimode fiber cable  It causes the input light pulse to spread.  Light Pulse consists of group of modes. The light energy is delayed with different amount along the fiber.

59. Graded Index Fiber Structure In graded index fiber, core refractive index decreases continuously with increasing radial distance r from center of fiber and constant in cladding Alpha defines the shape of the index profile As Alpha goes to infinity, above reduces to step index The index difference is

60. contd NA is more complex that step index fiber since it is function of position across the core Geometrical optics considerations show that light incident on fiber core at position r will propagate only if it within NA(r) Local numerical aperture is defined as And

61. contd Number of bound modes

62. Examples If a = 9.5 micron, find n2 in order to design a single mode fiber, if n1=1.465. Solution, The longer the wavelength, the larger refractive index difference is needed to maintain single mode condition, for a given fiber

63. Examples Compute the number of modes for a fiber whose core diameter is 50 micron. Assume that n1=1.48 and n2=1.46. Wavelength = 0.8 micron. Solution For large V, the total number of modes supported can be estimated as

64. Example What is the maximum core radius allowed for a glass fiber having n1=1.465 and n2=1.46 if the fiber is to support only one mode at wavelength of 1250nm. Solution